- #1
Yura
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- 0
this is the problem and I've worked for mostof it, but I am not sure I've got the right way to go about it, or if the ansewr i have found it right.
here, I am using the method for finding the vector equation
given the plane 2x + 6y + z = 6 intersects the paraboloid z = x^2 + y^2, find and name this curve of intersection.
__________________________________________
so far, i rearranged the plane equation to a function of z and let it equal te equation of the paraboloid.
this gave me:
x^2 + y^2 = 6 -2x - 6y
6 = x^2 + 2x + y^2 + 6y
6 = [x^2 + 2x + 1^2] + [y^2 + 6y +3^2] -10
16 = [x + 1]^2 + [y + 3]^2
4^2 = [x + 1]^2 + [y + 3]^2
this then gave me the projection downwards of the function, a circle. but the function should be an ellipse. (which can't be called a function y of x, because it fails the vertical test, not to clear on this)
to get the function of the curve of intersection, (here is where I am not sure about where I am going) i wanted x infunction of t and y in function of t (for parametric equations)
x = f(t)
y = g(t)
so i went:
rule: (cos(t))^2 + (sin(t))^2 = 1
[(x +1)/4]^2 + [(y + 3)/4]^2 = 1
(x +1)/4 = cos(t) => x + 1 = 4*cos(t) => x = 4*cos(t) - 1
(y + 3)/4 = sin(t) => y + 3 = 4*sin(t) => y = 4*sin(t) - 3
[(respectively): eqn, first eqn, second eqn]
[where o<= t<= 2*pi]
from,
4^2 = [x + 1]^2 + [y + 3]^2
=> 4^2 = [4*cos(t)]^2 + [4*sin(t)]^2
4^2 = 16*[cos(t)]^2 + 16*[sin(t)]^2
4^2 = 16*([cos(t)]^2 + [sin(t)]^2)
4^2 = 16*([cos(t)]^2 + [sin(t)]^2)
substituting second eqn x and y values into the plane:
z = 6 - 2x - 6y
=> z = 6 - 2*[4*cos(t) -1] - 6*[4*sin(t) - 3]
z = 6 - 8*cos(t) +2 - 24*sin(t) + 18
z = -8*cos(t) - 24*sin(t) + 26
this is what i have done so far (stop me if I am doing something completely wrong please ^^') but I'm a little lost in what the textbook is telling me
therefore the parametric equations for the equation
here is where everything goes iffy (if not already somewhere above).
when i use the parametric equation, because the projection is a circle the parametric equations would be:
x = 4*cos(t) - 1
y = 4*sin(t) - 3
z = -8*cos(t) - 24*sin(t) + 26
[where o<= t<= 2*pi]
so the corresponding vector equation is:
r(t) = [4*cos(t) - 1]i + [4*sin(t) - 3]j + [-8*cos(t) - 24*sin(t) + 26]k
[where o<= t<= 2*pi]
sorry if my explanation is unclear in some parts. I am not sure if this method gets me to the answer i want, but is there possibly another way to solve this problem? this was the only way i could think of.
the textbook didnt exactly teach the topic but there was an example that was similar to this problem I am working on so i was working a lot of this from that example (which was a really simple example and only had four lines for working).
here, I am using the method for finding the vector equation
given the plane 2x + 6y + z = 6 intersects the paraboloid z = x^2 + y^2, find and name this curve of intersection.
__________________________________________
so far, i rearranged the plane equation to a function of z and let it equal te equation of the paraboloid.
this gave me:
x^2 + y^2 = 6 -2x - 6y
6 = x^2 + 2x + y^2 + 6y
6 = [x^2 + 2x + 1^2] + [y^2 + 6y +3^2] -10
16 = [x + 1]^2 + [y + 3]^2
4^2 = [x + 1]^2 + [y + 3]^2
this then gave me the projection downwards of the function, a circle. but the function should be an ellipse. (which can't be called a function y of x, because it fails the vertical test, not to clear on this)
to get the function of the curve of intersection, (here is where I am not sure about where I am going) i wanted x infunction of t and y in function of t (for parametric equations)
x = f(t)
y = g(t)
so i went:
rule: (cos(t))^2 + (sin(t))^2 = 1
[(x +1)/4]^2 + [(y + 3)/4]^2 = 1
(x +1)/4 = cos(t) => x + 1 = 4*cos(t) => x = 4*cos(t) - 1
(y + 3)/4 = sin(t) => y + 3 = 4*sin(t) => y = 4*sin(t) - 3
[(respectively): eqn, first eqn, second eqn]
[where o<= t<= 2*pi]
from,
4^2 = [x + 1]^2 + [y + 3]^2
=> 4^2 = [4*cos(t)]^2 + [4*sin(t)]^2
4^2 = 16*[cos(t)]^2 + 16*[sin(t)]^2
4^2 = 16*([cos(t)]^2 + [sin(t)]^2)
4^2 = 16*([cos(t)]^2 + [sin(t)]^2)
substituting second eqn x and y values into the plane:
z = 6 - 2x - 6y
=> z = 6 - 2*[4*cos(t) -1] - 6*[4*sin(t) - 3]
z = 6 - 8*cos(t) +2 - 24*sin(t) + 18
z = -8*cos(t) - 24*sin(t) + 26
this is what i have done so far (stop me if I am doing something completely wrong please ^^') but I'm a little lost in what the textbook is telling me
therefore the parametric equations for the equation
here is where everything goes iffy (if not already somewhere above).
when i use the parametric equation, because the projection is a circle the parametric equations would be:
x = 4*cos(t) - 1
y = 4*sin(t) - 3
z = -8*cos(t) - 24*sin(t) + 26
[where o<= t<= 2*pi]
so the corresponding vector equation is:
r(t) = [4*cos(t) - 1]i + [4*sin(t) - 3]j + [-8*cos(t) - 24*sin(t) + 26]k
[where o<= t<= 2*pi]
sorry if my explanation is unclear in some parts. I am not sure if this method gets me to the answer i want, but is there possibly another way to solve this problem? this was the only way i could think of.
the textbook didnt exactly teach the topic but there was an example that was similar to this problem I am working on so i was working a lot of this from that example (which was a really simple example and only had four lines for working).